How does operator-overloading work? Step by step

Question

What are the components of operator-overloading? How does it work? What values does it calculate and return? I need someone who can answer me step by step the logic of this part of C++
- order of objects' definitions
- time of operator 's execution
- void constructor

Answer 1

please read this for all the information and after that if you have any question then comment here. I am here to assist you.

https://www.programiz.com/cpp-programming/operator-overloading

Answer 2

//sample code
#include <iostream>
using namespace std;

class Check
{
private:
int i;
public:
Check(): i(0) { }

Check operator ++()
{
Check temp;
++i;
temp.i = i;
return temp;
}

void Display()
{
cout << "i=" << i << endl;
}
};

int main()
{ Check obj, obj1;
obj.Display();
++obj;
obj.Display();
obj1 = ++obj;
obj1.Display();
return 0;
}

Answer 3

OrHy3 I did some change here for better understanding.

from prints you will get this-> var represents the object using which operator overloading method called and obj.var represents the object which we are passing.

please note the other change,

MyClass res = obj1+obj2;

Internally compiler will call like below,
MyClass res = obj1.operator+(obj2);

so now you can easily get way this->var has value of obj1 and obj.var has obj2 value.

we can also use internal code which compiler used to resolved operator overload function call.

#include <iostream>
using namespace std;

class MyClass {
public:
int var;
MyClass() { }
MyClass(int a)
: var(a) { }

MyClass operator+(MyClass &obj) {
MyClass res;
cout<<this->var<<endl;
cout<<obj.var<<endl;
res.var= this->var+obj.var;
return res;
}
};

int main() {
MyClass obj1(12), obj2(55);
MyClass res = obj1.operator+(obj2);

cout << res.var;
}

Answer 4

thanks

Answer 5

How can I use more than one object in an operation?

Answer 6

https://code.sololearn.com/cFQeE3OehCx9/?ref=app

Answer 7

https://code.sololearn.com/cH179plEBuKq/?ref=app

Answer 8

https://code.sololearn.com/cQgU53Wm0kU2/?ref=app

I don 't understand why in the operator function this->var and obj.var don 't have the same value

Answer 9

OrHy3 try to run above code and you will get the idea.

Answer 10

So you 're saying to me that the operator function is executed by obj1 and use obj2 as its method, right?

Answer 11

You can more then two object for operation by using grouping via parentheses.

example:
#include <iostream>
using namespace std;

class MyClass {
public:
int var;
MyClass() { }
MyClass(int a)
: var(a) { }

MyClass operator+(MyClass &obj) {
MyClass res;
res.var= this->var+obj.var;
return res;
}
};

int main() {
MyClass obj1(12), obj2(55), obj3(100);
MyClass res = (obj1 + obj2 )+ obj3;

cout << res.var;
}

Answer 12

thanks too much

Answer 13

no, not method obj2 as argument.
operator+ method is called for obj1 so this represents obj1. obj2 is passed in argument so obj.var represents obj2.

Answer 14

OrHy3 You are always wel come. Still if you have any questions then comment here. I will try my best to make it understandable.

Answer 15

ok, thanks

Answer 16

Just one question more: if there is more than one object, is the operator function called for the first object?

Answer 17

OrHy3 No not like that.
example:

(o1 + o2 ) + o3.

first parentheses should execute.
so ( o1 + o 2) , in this case operator method is called for o1 and o2 is passed as argument. returned object is stored in temporary object because + o3 is pending to execute.

now the expression becomes,
temp + o3. so operator method is called for temp and o3 is passed into argument.

Answer 18

I thought in (o1 + o2) the function is called for o1, not for o2

Answer 19

https://code.sololearn.com/cQgU53Wm0kU2/?ref=app

Why does this code need a void constructor to work?

Answer 20

OrHy3 check line no 12 and 20,which needs default constructor.

if you don't want to do that try like below,

#include <iostream>
using namespace std;

class MyClass {
public:
int var;
// MyClass() { }
MyClass(int a)
: var(a) { }

MyClass operator+(MyClass &obj) {
MyClass res(0);
res.var= this->var+obj.var;
return res;
}
};

int main() {
MyClass obj1(12), obj2(55);
MyClass res (obj1+obj2);

cout << res.var;
}